For $c_{1},c_{2}\in \mathbb{H}:=\{Im(z)>0\}$ I want to compute the following integral or prove it doesn't exist:
$$\int_{\mathbb{R}\times \mathbb{R}^{+}}\int_{\mathbb{R}\times \mathbb{R}^{+}}e^{-|w-c_{1}|^2-|u-c_{2}|^2}\frac{1}{w_{1}+iw_{2}-u_{1}-iu_{2}}dw_{1}dw_{2}du_{1}du_{2},$$
where $w_{1},u_{1}\in \mathbb{R}$,$w_{2},u_{2}\in \mathbb{R}^{+}$. Please only hints to help me practice.
Q1 Is there a way to turn this into a contour integral?
It was on mathstackexchange for a while but I didn't get any help, so I deleted it.
Q2 Is it possible that it equals $c\cdot \frac{1}{c_{1}-c_{2}}$?
How would I go about disproving this? This guess came from treating it as a Gaussian integral, doing Laplace method and a power series approximation. One reason I think it is not true is because as we take $c_{1}-c_{2}\to 0$ the right hand side blows up, whereas the left is still bounded from attempt (1).
Attempts
1)Existence of integral. Around a neighbourhood of zero we can bound the integral by
$$c\int_{0}^{\varepsilon}\int_{0}^{\varepsilon} \frac{r\rho}{|r-\rho|}dr d\rho,$$
for $w=re^{i\theta},u=\rho e^{i\phi}$ and this can be computed. Outside that neighbourhood, we have the bound
$$c'\int_{\varepsilon}^{\infty}\int_{\varepsilon}^{\infty}e^{-r^{2}-\rho^{2}} \frac{r\rho}{|r-\rho|}dr d\rho,$$
which is also bounded.
2)Somehow turning this into a contour integral.
$$\int e^{-|w-c_{1}|^{2}-|u-c_{2}|^{2}} \frac{r\rho}{re^{2i\theta}-e^{i\theta}\rho e^{i\phi}}e^{i\theta}d\theta dr $$
$$\int_{r=0}^{\infty}\int_{|w|=r,arg(w)\in [0,\pi]} e^{-|w-c_{1}|^{2}-|u-c_{2}|^{2}} \frac{|w|\rho}{w(w-\rho e^{i\phi})}dw dr$$
and so considering the contour integral over the upper semicircle indented to include pole 0. However, the gaussian is not holomorphic and so we cannot apply the residue theorem.
3)Trying to use the rectangular contour used for the gaussian: $[-R,R]\cup [-R+\tau,R+\tau\cup [-R,-R+\tau]\cup [R,R+\tau]$. However, this required a certain periodic integrand, which we don't have i.e. $f(w):=e^{-w^{2}}\frac{|w|}{w(w-u)}$ with the property $f(w)-f(w+\tau)=e^{-w^{2}}\frac{|w|}{w(w-u)}$.
4)Somehow turning this into a contour integral via Green's theorem: it would require writing $Q_{x}-P_{y}=e^{-|(x,y)|^{2}}\frac{1}{x+iy-u)}$. All we have is $[\partial_{x}-i\partial_{y}]Log(w-u)$.
5)We do change of variables $\widetilde{w}=w-c_{1},\widetilde{u}=u-c_{2}$ and polar coordinates we obtain:
$$\int_{\mathbb{R}\times [-iIm(c_{2}),\infty]}\int_{\mathbb{R}\times [-iIm(c_{1}),\infty]}e^{-|w|^2-|u|^2}\frac{1}{w-u+c_{1}+c_{2}}dw_{1}dw_{2}du_{1}du_{2}.$$
We split the domain $[-iIm(c_{2}),\infty]=[-iIm(c_{2}),0]\cup (0,\infty]$ in order to use polar coordinates; we start with the easier part:
$$\int_{[0,\infty]\times [0,\pi]}\int_{[0,\infty]\times [0,\pi]}e^{-r^2-\rho^2}\frac{r \rho}{re^{i\theta}-\rho e^{i\phi}+c}drd\theta d\rho d\phi.$$
Now this we can study one iterated integral at a time:
$$\int_{[0,\infty]}e^{-r^2}r\int_{ [0,\pi]}\frac{1 }{re^{i\theta}-u+c}d\theta dr.$$
My problem is studying the other part $[-iIm(c_{2}),0]$, which I will post as I find things.
